5.4 Comparison tests (Page 3/7)

Calculus volume 2 Page 3 / 7

Limit comparison test

Let $a_{n}, b_{n} \geq 0$ for all $n \geq 1 .$

If $lim_{n \to \infty} a_{n} / b_{n} = L \neq 0,$ then $\sum_{n = 1}^{\infty} a_{n}$ and $\sum_{n = 1}^{\infty} b_{n}$ both converge or both diverge.
If $lim_{n \to \infty} a_{n} / b_{n} = 0$ and $\sum_{n = 1}^{\infty} b_{n}$ converges, then $\sum_{n = 1}^{\infty} a_{n}$ converges.
If $lim_{n \to \infty} a_{n} / b_{n} = \infty$ and $\sum_{n = 1}^{\infty} b_{n}$ diverges, then $\sum_{n = 1}^{\infty} a_{n}$ diverges.

Note that if $a_{n} / b_{n} \to 0$ and $\sum_{n = 1}^{\infty} b_{n}$ diverges, the limit comparison test gives no information. Similarly, if $a_{n} / b_{n} \to \infty$ and $\sum_{n = 1}^{\infty} b_{n}$ converges, the test also provides no information. For example, consider the two series $\sum_{n = 1}^{\infty} 1 / \sqrt{n}$ and $\sum_{n = 1}^{\infty} 1 / n^{2} .$ These series are both p -series with $p = 1 / 2$ and $p = 2,$ respectively. Since $p = 1 / 2 > 1,$ the series $\sum_{n = 1}^{\infty} 1 / \sqrt{n}$ diverges. On the other hand, since $p = 2 < 1,$ the series $\sum_{n = 1}^{\infty} 1 / n^{2}$ converges. However, suppose we attempted to apply the limit comparison test, using the convergent $p - series$ $\sum_{n = 1}^{\infty} 1 / n^{3}$ as our comparison series. First, we see that

\frac{1 / \sqrt{n}}{1 / n^{3}} = \frac{n^{3}}{\sqrt{n}} = n^{5 / 2} \to \infty as n \to \infty .

Similarly, we see that

\frac{1 / n^{2}}{1 / n^{3}} = n \to \infty as n \to \infty .

Therefore, if $a_{n} / b_{n} \to \infty$ when $\sum_{n = 1}^{\infty} b_{n}$ converges, we do not gain any information on the convergence or divergence of $\sum_{n = 1}^{\infty} a_{n} .$

Using the limit comparison test

For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.

$\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n} + 1}$
$\sum_{n = 1}^{\infty} \frac{2^{n} + 1}{3^{n}}$
$\sum_{n = 1}^{\infty} \frac{ln (n)}{n^{2}}$

Compare this series to $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}} .$ Calculate

$lim_{n \to \infty} \frac{1 / (\sqrt{n} + 1)}{1 / \sqrt{n}} = lim_{n \to \infty} \begin{matrix} \frac{\sqrt{n}}{\sqrt{n} + 1} \end{matrix} = lim_{n \to \infty} \frac{1 / \sqrt{n}}{1 + 1 / \sqrt{n}} = 1 .$
By the limit comparison test, since $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}}$ diverges, then $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n} + 1}$ diverges.
Compare this series to $\sum_{n = 1}^{\infty} {(\frac{2}{3})}^{n} .$ We see that

$lim_{n \to \infty} \frac{(2^{n} + 1) / 3^{n}}{2^{n} / 3^{n}} = lim_{n \to \infty} \frac{2^{n} + 1}{3^{n}} \cdot \frac{3^{n}}{2^{n}} = lim_{n \to \infty} \frac{2^{n} + 1}{2^{n}} = lim_{n \to \infty} [1 + {(\frac{1}{2})}^{n}] = 1 .$

Therefore,

$lim_{n \to \infty} \frac{(2^{n} + 1) / 3^{n}}{2^{n} / 3^{n}} = 1 .$

Since $\sum_{n = 1}^{\infty} {(\frac{2}{3})}^{n}$ converges, we conclude that $\sum_{n = 1}^{\infty} \frac{2^{n} + 1}{3^{n}}$ converges.
Since $ln n < n,$ compare with $\sum_{n = 1}^{\infty} \frac{1}{n} .$ We see that

$lim_{n \to \infty} \frac{ln n / n^{2}}{1 / n} = lim_{n \to \infty} \frac{ln n}{n^{2}} \cdot \frac{n}{1} = lim_{n \to \infty} \frac{ln n}{n} .$

In order to evaluate $lim_{n \to \infty} ln n / n,$ evaluate the limit as $x \to \infty$ of the real-valued function $ln (x) / x .$ These two limits are equal, and making this change allows us to use L’Hôpital’s rule. We obtain

$lim_{x \to \infty} \frac{ln x}{x} = lim_{x \to \infty} \frac{1}{x} = 0 .$

Therefore, $lim_{n \to \infty} ln n / n = 0,$ and, consequently,

$lim_{n \to \infty} \frac{ln n / n^{2}}{1 / n} = 0 .$

Since the limit is $0$ but $\sum_{n = 1}^{\infty} \frac{1}{n}$ diverges, the limit comparison test does not provide any information.
Compare with $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ instead. In this case,

$lim_{n \to \infty} \frac{ln n / n^{2}}{1 / n^{2}} = lim_{n \to \infty} \frac{ln n}{n^{2}} \cdot \frac{n^{2}}{1} = lim_{n \to \infty} ln n = \infty .$

Since the limit is $\infty$ but $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ converges, the test still does not provide any information.
So now we try a series between the two we already tried. Choosing the series $\sum_{n = 1}^{\infty} \frac{1}{n^{3 / 2}},$ we see that

$lim_{n \to \infty} \frac{ln n / n^{2}}{1 / n^{3 / 2}} = lim_{n \to \infty} \frac{ln n}{n^{2}} \cdot \frac{n^{3 / 2}}{1} = lim_{n \to \infty} \frac{ln n}{\sqrt{n}} .$

As above, in order to evaluate $lim_{n \to \infty} ln n / \sqrt{n},$ evaluate the limit as $x \to \infty$ of the real-valued function $ln x / \sqrt{x} .$ Using L’Hôpital’s rule,

$lim_{x \to \infty} \frac{ln x}{\sqrt{x}} = lim_{x \to \infty} \frac{2 \sqrt{x}}{x} = lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0 .$

Since the limit is $0$ and $\sum_{n = 1}^{\infty} \frac{1}{n^{3 / 2}}$ converges, we can conclude that $\sum_{n = 1}^{\infty} \frac{ln n}{n^{2}}$ converges.

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Use the limit comparison test to determine whether the series $\sum_{n = 1}^{\infty} \frac{5^{n}}{3^{n} + 2}$ converges or diverges.

The series diverges.

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Key concepts

The comparison tests are used to determine convergence or divergence of series with positive terms.
When using the comparison tests, a series $\sum_{n = 1}^{\infty} a_{n}$ is often compared to a geometric or p -series.

Use the comparison test to determine whether the following series converge.

$\sum_{n = 1}^{\infty} a_{n}$ where $a_{n} = \frac{2}{n (n + 1)}$

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$\sum_{n = 1}^{\infty} a_{n}$ where $a_{n} = \frac{1}{n (n + 1 / 2)}$

Converges by comparison with $1 / n^{2} .$

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Questions & Answers

anyone know any internet site where one can find nanotechnology papers?

Damian Reply

research.net

kanaga

Introduction about quantum dots in nanotechnology

Praveena Reply

what does nano mean?

Anassong Reply

nano basically means 10^(-9). nanometer is a unit to measure length.

Bharti

do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?

Damian Reply

absolutely yes

Daniel

how to know photocatalytic properties of tio2 nanoparticles...what to do now

Akash Reply

it is a goid question and i want to know the answer as well

Maciej

characteristics of micro business

Abigail

for teaching engĺish at school how nano technology help us

Anassong

Do somebody tell me a best nano engineering book for beginners?

s. Reply

there is no specific books for beginners but there is book called principle of nanotechnology

NANO

what is fullerene does it is used to make bukky balls

Devang Reply

are you nano engineer ?

fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.

Tarell

what is the actual application of fullerenes nowadays?

Damian

That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.

Tarell

what is the Synthesis, properties,and applications of carbon nano chemistry

Abhijith Reply

Mostly, they use nano carbon for electronics and for materials to be strengthened.

Virgil

is Bucky paper clear?

CYNTHIA

carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc

NANO

so some one know about replacing silicon atom with phosphorous in semiconductors device?

s. Reply

Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.

Harper

Do you know which machine is used to that process?

how to fabricate graphene ink ?

SUYASH Reply

for screen printed electrodes ?

SUYASH

What is lattice structure?

s. Reply

of graphene you mean?

Ebrahim

or in general

Ebrahim

in general

Graphene has a hexagonal structure

tahir

On having this app for quite a bit time, Haven't realised there's a chat room in it.

Cied

what is biological synthesis of nanoparticles

Sanket Reply

what's the easiest and fastest way to the synthesize AgNP?

Damian Reply

China

Cied

types of nano material

abeetha Reply

I start with an easy one. carbon nanotubes woven into a long filament like a string

Porter

many many of nanotubes

Porter

what is the k.e before it land

Yasmin

what is the function of carbon nanotubes?

Cesar

I'm interested in nanotube

Uday

what is nanomaterials and their applications of sensors.

Ramkumar Reply

Difference between extinct and extici spicies

Amanpreet Reply

Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?

Abdul Reply

You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?

Abdul

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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